Bernoulli and Continuity Equations with Venturimeter Worked Examples

Two equations solve a surprising number of fluid flow problems. The continuity equation says that mass is conserved, so a fluid speeds up where a pipe narrows. Bernoulli's equation says that energy is conserved along a streamline, so where the fluid speeds up its pressure must fall. Together they explain flow meters, carburettors, spray nozzles and the water jet from a tank. This article states both equations, explains their assumptions and solves three practical problems.

The Equations

Continuity (incompressible):   Q = A1 * V1 = A2 * V2

Bernoulli (per unit weight, in metres of fluid):
   p1/(rho g) + V1^2/(2g) + z1 = p2/(rho g) + V2^2/(2g) + z2

   p/(rho g)  = pressure head
   V^2/(2g)   = velocity head
   z          = datum (elevation) head

With losses and a pump:
   H1 + H_pump = H2 + h_loss

Bernoulli's equation in its basic form assumes steady flow, incompressible fluid, no friction and no shaft work between the two points, and applies along a single streamline. Water in short pipes and air at low speed (below roughly 0.3 times the speed of sound) satisfy these assumptions well enough for first estimates.

Quantity	Symbol	Unit	Water value used here
Density	rho	kg/m^3	1000
Gravity	g	m/s^2	9.81
Discharge	Q	m^3/s	1 L/s = 0.001 m^3/s
Specific gravity of mercury	S_m	none	13.6

Example 1: Pressure Drop in a Pipe Reducer

Water flows at 10 L/s through a horizontal pipe that reduces from 100 mm to 50 mm diameter. Find the velocities and the pressure difference between the two sections, ignoring losses.

A1 = pi/4 * 0.1^2   = 7.854e-3 m^2
A2 = pi/4 * 0.05^2  = 1.963e-3 m^2

V1 = 0.01 / 7.854e-3 = 1.27 m/s
V2 = 0.01 / 1.963e-3 = 5.09 m/s

z1 = z2, so  p1 - p2 = (rho/2)(V2^2 - V1^2)
            = 500 * (25.94 - 1.62)
            = 12,160 Pa = 12.2 kPa

Halving the diameter quarters the area and so increases velocity four times. Because velocity head depends on the square of velocity, the pressure drop grows rapidly with any further reduction.

Example 2: Jet from a Tank (Torricelli)

Water leaves a small hole 5 m below the free surface of a large open tank. Take point 1 on the free surface and point 2 in the jet. Both are at atmospheric pressure, and the surface velocity is nearly zero because the tank is large.

z1 - z2 = h = 5 m
V2 = sqrt(2 g h) = sqrt(2 * 9.81 * 5) = 9.90 m/s

The real jet is slower because of friction, and narrower because the stream contracts after leaving the hole (vena contracta). For a sharp-edged orifice the coefficient of discharge is commonly around 0.6, so actual flow is roughly 60 percent of the ideal value.

Example 3: Venturimeter with Mercury Manometer

A venturimeter measures flow by creating a pressure drop at a throat. Combining continuity and Bernoulli between the inlet (area A1) and the throat (area A2) gives the discharge formula below. The differential manometer converts the mercury column reading x into an equivalent head of water h.

Q = Cd * A1 * A2 * sqrt(2 g h) / sqrt(A1^2 - A2^2)
h = x * (S_m / S_w - 1)

A 200 mm by 100 mm venturimeter in a horizontal water pipe shows a mercury manometer reading of 200 mm. Take Cd = 0.98.

h  = 0.2 * (13.6 - 1) = 2.52 m of water
A1 = pi/4 * 0.2^2 = 0.031416 m^2
A2 = pi/4 * 0.1^2 = 0.007854 m^2

sqrt(2 g h)          = sqrt(2 * 9.81 * 2.52) = 7.032 m/s
sqrt(A1^2 - A2^2)    = 0.030418 m^2
A1 * A2              = 2.4674e-4 m^4

Q = 0.98 * 2.4674e-4 * 7.032 / 0.030418
  = 0.0559 m^3/s = 55.9 L/s

Notice that the formula needs only the pressure difference, not the absolute pressures. That is why a single differential manometer or a differential pressure transmitter connected across the inlet and throat tappings is enough to measure flow. In an inclined venturimeter the elevation difference between the tappings cancels out of the manometer reading as well, so the same formula applies.

A related instrument: the pitot tube

A pitot tube faces into the flow and brings the fluid at its tip to rest. The rise in pressure at the tip over the static pressure equals the velocity head, so V = sqrt(2 g h), where h is the difference between the stagnation and static heads. For example, a difference of 0.2 m of water gives a local velocity of sqrt(2 x 9.81 x 0.2) = 1.98 m/s. Pitot-static tubes on aircraft use the same principle to measure airspeed.

A venturimeter has a high coefficient of discharge, typically 0.95 to 0.99, because its gradual converging and diverging cones avoid flow separation. An orifice plate is cheaper and shorter but has a much lower coefficient and a larger permanent pressure loss.

Common Mistakes

Mixing litres and cubic metres. 1 L/s is 0.001 m^3/s.
Using the mercury reading directly as head of water instead of multiplying by (13.6 - 1).
Applying Bernoulli across a pump or turbine without adding the shaft work term.
Ignoring elevation when the pipe is inclined; z1 and z2 must both be measured from the same datum.
Expecting ideal Bernoulli results in long pipes, where friction losses must be added using the Darcy-Weisbach equation.

Summary

Continuity links area and velocity; Bernoulli trades pressure, velocity and elevation heads along a streamline. Combining them lets you predict pressure changes in reducers, jet velocities from tanks and flow rates from venturimeter readings. Once friction losses and pump head are added, the same energy equation becomes the basis for pump selection, covered in the next article.